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A jet of water squirts out horizontally from a
hole on the side of the tank as shown below.
If the hole has a diameter of 3.57 mm , what
is the height of the water above the hole in
the tank?
Answer in units of cm.

A jet of water squirts out horizontally from a hole on the side of the tank as shown below If the hole has a diameter of 357 mm what is the height of the water class=

Respuesta :

elcharly64

Answer:

h=31.6 cm

Explanation:

Conservation of Energy in Fluids

Let's first take care of the motion magnitudes. We know the height and horizontal distance the jet of water has. It's similar to a horizontal launch of an object at a speed vo. The horizontal distance traveled is

[tex]x=v_o.t[/tex]

And the vertical distance is

[tex]\displaystyle y=\frac{gt^2}{2}[/tex]

From this last equation, we solve for t

[tex]\displaystyle t=\sqrt{\frac{2y}{g}}[/tex]

[tex]\displaystyle t=\sqrt{\frac{2\times 1.36}{9,8}[/tex]

[tex]t=0.278\ sec[/tex]

Now we find vo

[tex]\displaystyle v_o=\frac{x}{t}[/tex]

[tex]\displaystyle v_o=\frac{0.692}{0.278}[/tex]

[tex]v_o=2.49\ m/s[/tex]

Now we can set up the conditions for the conservation of energy for the fluid, assuming ideal conditions (no friction, compression, etc) we can state

[tex]\displaystyle \rho g h=\frac{\rho v_o^2}{2}[/tex]

solving for h

[tex]\displaystyle h=\frac{v_o^2}{2g}[/tex]

[tex]\displaystyle h=\frac{2.49^2}{2\times 9.8}[/tex]

[tex]h=0.316\ m[/tex]

[tex]\boxed{h=31.6\ cm}[/tex]

The velocity of the flowing fluid out of the opening is given by Torricelli's law.

  • The height of the water above the hole in the tank is approximately 8.803 cm.

Reasons:

The given diameter of the hole, d ≈ 3.57 mm (small hole)

By Torricelli's law, we have;

v = √(2·g·h)

By Newton's Laws of motion, we have;

[tex]\mathrm{The \ time \ it \ takes \ the \ water \ to \ reach \ the \ ground, \ t } = \mathbf{\sqrt{\dfrac{2 \cdot y}{g}} }[/tex]

Therefore;

[tex]\mathrm{Horizontal \ velocity \ of \ the \ water, \ v} = \mathbf{ \dfrac{x}{ \sqrt{\dfrac{2 \cdot y}{g} }}}[/tex]

[tex]\mathrm{\sqrt{2 \cdot g \cdot h} } = \dfrac{x}{ \sqrt{\dfrac{2 \cdot y}{g} }}[/tex]

Making, h the subject, of the above equation, gives;

[tex]h =\dfrac{x^2 \cdot g}{2 \cdot y\cdot 2 \cdot g} = \mathbf{\dfrac{x^2 }{2 \cdot y\cdot 2 }} = \dfrac{0.692^2}{2 \times 1.36\times 2} = 8.803\times 10 ^{-2}[/tex]

The value for the height above the hole, h ≈ 8.803 × 10⁻² m = 8.803 cm.

Learn more here:

https://brainly.com/question/13091900

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